int cos ^{3} x cdot e^{log (sin x)} d x is eq | vedclass

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Question

(B) Let $I = \int \cos ^{3} x \cdot e^{\log (\sin x)} d x$. Since $e^{\log (\sin x)} = \sin x$,the integral becomes $I = \int \cos ^{3} x \sin x d x$.

Vedclass · Accepted Answer

$-\frac{\cos ^{4} x}{4}+c$

Vedclass · Answer

(B) Let $I = \int \cos ^{3} x \cdot e^{\log (\sin x)} d x$. Since $e^{\log (\sin x)} = \sin x$,the integral becomes $I = \int \cos ^{3} x \sin x d x$. Let $u = \cos x$. Then $du = -\sin x d x$,which implies $\sin x d x = -du$. Substituting these into the integral,we get $I = \int u^{3} (-du) = -\int u^{3} du$. Integrating with respect to $u$,we get $I = -\frac{u^{4}}{4} + c$. Substituting back $u = \cos x$,we obtain $I = -\frac{\cos ^{4} x}{4} + c$.

$\int \cos ^{3} x \cdot e^{\log (\sin x)} d x$ is equal to

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